 So, in the previous lecture we have started looking at the algebraic and the geometric multiplicity of an operator, right and we saw that there are some interesting features such as if you were to transform this operator or look at this operator subject to a new basis at least the geometric and the algebraic multiplicities do not change. In fact, we have also seen that the characteristic polynomial also remains invariant under a change of basis, right. So, today we are going to probe a little deeper into this idea of geometric multiplicity and subsequently we are going to come up with a condition for diagonalizability of a linear operator, right. So, we have seen that of pivotal importance in our studies has been this particular subspace, right where a is of course the operator or you can treat it as a matrix as I said we will not distinguish them too often here, right. So, this is the subspace that we have actually given a name we have called it W i and it is precisely the dimension of W i perhaps I denoted it by the variable l i if not mistaken is that what it was, right. So, this is what we call the geometric multiplicity of lambda i which is an eigenvalue of a, right. So, today we are going to see an interesting property of this W i, right. So, consider lambda 1, lambda 2 till lambda k to be distinct eigenvalues of a, all right. Then look at W which we now define as W 1 plus W 2 plus so on till W k we understand what this means it is a sum of subspaces we are familiar with this object, right. Of course, our V is finite dimensional again whenever we studying eigenvalues and eigenvectors V is finite dimensional. We have said already that if you have infinite dimensional vector spaces the existence of eigenvalues cannot be guaranteed. So, we will do well to keep that in mind. The point that is going to be made here is very interesting which is suppose B i is a basis for W i, right. So, W i's are of course these objects, right. Then the union of these basis sets is a basis for W, right. So, that is our claim that if you look at these subspaces so defined, yeah corresponding to distinct eigenvalues lambda 1 through lambda k no two of which are equal, right. And you look at those subspaces and because these are finite dimensional you will always have basis for these subspaces. So, you take the basis of W 1, take the basis vectors of W 2 and take the union of this set of basis vectors for each of those subspaces that will give you a basis for W, right. Generally the above is a direct sum. So, at this point I have not claimed that this is a direct sum it is just a sum of subspaces. But I am now claiming that this is also going to be a direct sum, alright. What is it that you immediately understand when we say something is a direct sum? The intersections of any two of those, yeah, is going to be trivial except from for the 0 vector nothing else can be in common between these objects. So, that is the whole claim, okay. So, if you want to prove this we will of course constitute in showing two things that is the first part the part prior to this direct sum will involve showing two things. What are those two things? The fact that this set this is a set, right. This is a set it is a union of multiple sets. So, it is a set the fact that this set that I box now is a linearly independent set and that this is a generating set for W. If I show that then I will be done with the first part, right. So, first I will show that this is indeed a generating set for any vector in W. That should be pretty obvious actually the part that about this being a generating set is not very hard to see. So, consider W belonging to W, alright. So, if W belongs to this subspace W which is nothing but the sum of these subspaces I can write little W as W1 plus W2 plus till Wk where Wi belongs to the subspace Wi I going from 1 through k. After all that is what it means for something to belong to the sum of subspaces. Now, where is this object coming from then W1 this object from W2. What do we know to be a basis for W1? The subspace W1 I mean its objects in B1, right. So, let elements in Bi be given by. So, remember again because it is finite dimensional it will have a termination point you cannot have infinite number of objects in this basis sets, right. So, let elements in Bi be given by Bi is equal to Wi1, Wi2 until Wi let us say what should I what variable should I use Mi which means that the dimension of Wi is little Mi, yeah, right. No questions about this alright which means that W can be written as summation alpha ij Wij of course there will be a double summation or maybe I will just open up the bracket and help you see this easily, okay. So, let us just say summation W1j alpha 1j going from 1 through M1 plus summation alpha 2j W2jj going from 1 through M2 until alpha what is the number kj Wkjj going from 1 through Mk, clear. These are objects in the basis B1 these are objects linear combination of objects in the basis B2 likewise these are linear combination of objects in the basis Bk. So, this then obviously belongs to the span of W11, W12 until W1M1, W21, W22 until W2M2 likewise until Wk1, Wk2 until WkMk, right. I could have used a double summation I just felt that this helps in better visualization which is nothing but in the span of what the union of Bi's isn't it, yeah. This is the union of Bi's I have just stitched together all the elements in individual basis sets and I have gotten this huge yeah Mk times k this large set right. So, this definitely is a generating set implies union Bk sorry union Bi I should write is a spanning set for W. So, the spanning part of this candidate has been verified now we need to ensure that this set this large set that we have here is also linearly independent right. So, I am going to erase this part. So, consider I am going to just use the same notation alright is going to be very handy and you will see in a moment why summation yes no we need to show that this set of vectors any vector that you have from a subspace belongs to the span yeah. So, that means this set actually does not contain anything more than W no, no, no. So, you might have some redundancies. So, for example, a basis for a original vector space is also a generating set for a subspace or proper subspace of the vector space you might just assign a weight 0 to those vectors in the basis which do not belong to that subspace. So, you might have redundancies, but of course, in that case it does not make sense because the objects themselves must also come from that subspace. So, in general of course, we do not encounter such situations. So, that is only trivially obvious the fact is that this is what we were required to show that this is generating set and we have done that. So, next part is to show that it is also linearly independent if it is linearly independent I put forth to you that this sum that I have written here yeah. So, that is why I did not I actually thought better of this than to use the double summation here because I am going to use this. So, if I now posit that this has to be linearly independent I have to first check that when this sum becomes equal to 0 it is only possible if each of these alpha 1 j alpha 1 1 alpha 1 2 till alpha 1 m 1 alpha 2 1 alpha 2 2 till alpha 2 m 2 and so on till alpha k 1 alpha k 2 so on till alpha k m k all of them have to vanish. If that is so only then I will be able to conclude that it is linearly independent. So, consider the same sum here except that the posing of this is a little different now. So, I am going to still use the let me use betas. So, I want to confuse you. So, let us say this is beta 1 j w 1 j plus summation j running from 1 through m 2 beta 2 j w 2 j plus so on till summation j running from 1 through m k beta k j w k j is equal to 0. Now, there is something very interesting that will happen forget about this part just focus on this we are claiming that this sum is 0 when is it possible? Now, if I just focus on this object where does this come from w 1 if I focus on this object this comes from w 2 if I focus on this object this comes from w k similarly. So, this means that no matter what this combination is after all at the end of the day this is just sum w 1 plus w 2 till w k that is vanishing right. So, this is my w 1 this is my w 2 and this is my w k. So, if this has to vanish then I must have a bunch of vectors each coming from those different kernels or different w i's whose sum must vanish, but what are these w i's after all? They are in the kernel of a minus lambda i i what sort of vectors live inside the kernel of a minus lambda i i the eigenvectors is not it. So, w 1 is an eigenvector corresponding to lambda 1 w 2 is an eigenvector corresponding to lambda 2 dot until w k is an eigenvector corresponding to lambda k and these are distinct eigenvalues. What do we know about eigenvectors corresponding to distinct eigenvalues? Lineal independent. Lineal independent and this is definitely a legit linear combination of linearly independent fellows. So, they cannot be 0 do you see they cannot be 0 unless all are zeros themselves. So, they are not even eigenvectors if they are eigenvectors then this is not possible. So, this is not possible unless w i is equal to 0 for i is equal to 1 to until k please ask if this is not clear this is the crux of the proof essentially please ask if this is not clear do not hesitate I will just repeat just in case you are missing the point here. The point is that this sum is just a basic test for linear independence right if it is linearly independent then each of these objects must vanish. So, I just put a linear combination of these fellows and equate it to 0 and check whether it is possible for any non-trivial linear combination straightforward very basics. I observe that this sum the first term of this comes from w 1. So, let us label it as w 1 the second term comes from the subspace w 2 let us call it little w 2. So, on the k th term comes from the subspace w k let us label it little w k and then I equate the sum to 0. The next observation is the crucial bit where I observe that after all objects inside the subspace w i are nothing, but eigenvectors corresponding to lambda i and the lambda i is by my very premise are all distinct and earlier much earlier in this course maybe some 4 5 lectures back we have proved that eigenvectors corresponding to distinct eigenvalues must be linearly independent and therefore, any linear combination you take of those eigenvectors that must be non-zero right any non-trivial linear combination. So, this is one such non-trivial linear combination because each of the coefficients is 1. So, what are we possibly missing the only possibility is that these fellows cannot be legitimate eigenvectors. How can we ensure that they are not legitimate eigenvectors? Well there is this one deal which says that the eigenvectors must be non-zero anything is inside any 0 vector is inside the kernel of any operator. So, only when the vector is non-zero non-trivial then it makes sense to define it as an eigenvector otherwise 0 will be the eigenvector for everything. No sense right. So, therefore, the only way that these fellows fail to be eigenvectors are if each of them individually is 0. If any one of them fails to be 0 then we cannot have this situation because then it is an eigenvector and linear combination of eigenvectors corresponding to distinct eigenvalues can never vanish. Unless the linear combination is trivial which it clearly is not is that clear. Now, if each individual w i is 0 that means each of these individual sums becomes 0 which means that summation beta i j j going from 1 through m i w i j is equal to 0. But these are after all what these are elements in the basis b i and a basis by definition is linearly independent. So, therefore, I arrive at the conclusion that. So, this is true for all i yeah. So, beta i j is equal to 0 for all i belonging to the set 1 2 until k and for all j belonging to the set 1 2 3 until m i. But that is essentially the definition of linear independence of these fellows which means that the union of the basis sets is also a linearly independent set right. So, let me erase this part now that means it is a linearly independent set it is a generating set. Of course, therefore, by putting together those 2 things now for the final bit which is when we have to show that it is a direct sum right. So, consider w belonging to intersection of w i and w j for i not equal to j which means that a minus lambda i i acting on w is equal to 0 a minus lambda j i acting on w is equal to 0. You take the difference and you are led to concluding that lambda j minus lambda i times w is equal to 0. But lambda j minus lambda i is clearly not equal to 0 because they are distinct eigenvalues yeah. So, therefore, w is equal to 0. So, anytime you have something in the intersection of any 2 of those subspaces it must be only the 0 vector and nothing more than that and therefore, indeed that was a direct sum right and you can also check by the dimensions now. Individual dimensions if they are m i then the overall dimension is summation m i that also tallies with our understanding of dimensions and how they work out over direct sums right. So, this is an important result which is going to pave the way for now our big result which is going to talk about diagonalizability of an operator. So, this part is clear if it is then I will probably it is that part and write our next result. So, for a which is an operator from v to itself of course, finite dimensional vector space let us just put it right there even if it is obvious ok. The following are equivalent 1 a is diagonalizable to algebraic multiplicity of lambda i is equal to dimension w i for all i ok lambda i obviously denotes the i th eigen value of a. So, of course, all that is implied here alright and third a has n. So, of course, finite dimensional let us just say the dimension is n that will make life easier a has n linearly independent eigen vectors. So, we are not talking about distinct eigen values or anything as it turns out we do not need distinct eigen values one clear cut example as I have mentioned earlier is the identity matrix all its eigen values are one they are all repeated the extreme case and yet it is a diagonal matrix. So, obviously, you do not require all n eigen values to be distinct it is a sufficient condition for diagonalizability as we have seen. Now, this is exactly or precisely the condition ok if you want diagonalizability then this is by the way what the geometric multiplicity of lambda i you may also write it as such right. So, I might have of course, defined the characteristic polynomial and shown you all that product form and all, but instead because I have already taken the trouble of defining the algebraic multiplicity I am just going to use that notation here yeah. This of course, is the definition of the geometric multiplicity this part right ok. What does it mean when we say that an operator is diagonalizable see the main crux of this will again be down to understanding what these statements actually mean once you see what they mean things will become very obvious. So, let us first try and understand or parse these statements. When we say that a is diagonalizable what we are essentially saying is that there is a basis in v corresponding to which the representation of the operator a turns out to be a diagonal matrix that is the meaning of diagonalizable. Remember right at the beginning of this course when we were talking about the preliminary goals of this course one was to solve a x is equal to b and the second was this p inverse a p getting it down to a diagonal form what is the p inverse a p now having done so many so much of this course having covered so much of this we have a better understanding that p inverse a p is just the representation of a subject to some change of basis some basis. So, if we can say that p inverse a p is equal to some diagonal matrix it means a is diagonalizable or in the language of basis we understand that there exists some basis subject to the choice of which the operator looks completely like a decoupled and we have seen the benefits of this in particularly in solving differential equations right you have nth order or nth degree differential equation it reduces to just solving n first order differential equations that is very simple is it not right. So, let us try and prove this what do you mean by unique basis I mean you take any vector you scale it up and down it is still an eigenvector you take any vector in the span of that particular vector right that is also an eigenvector is it not if 1 0 is an eigenvector of a particular 2 by 2 matrix then so is 1000 0 1 0 1000 0 1 billion 0 or 10 to the power minus 30 0 all of them are eigenvectors. So, I might choose a basis you might choose another basis you might scale up those matrices in different order in different manner there is no claim for uniqueness of eigenvectors eigenvalues sure unless you put more restriction that the eigenvector has to be of unit norm or something of such sort you will never be able to get it down to some unique object even so you might have multiple eigenvectors for the same eigenvalue for example, for the identity yeah you take any of the standard basis elements they are its eigenvectors but the funny thing is you take any vector in the Euclidean space it is an eigenvector so if you are already starting with the identity matrix you take any basis so such a basis is obviously not going to be unique right. So, there is no claim for uniqueness we will get into this question maybe after covering quite a bit we will get into this when something is diagonalizable and when things are not diagonalizable and so on and so forth and when they are not diagonalizable we will see the best we can do is Jordan form and then we will see why it makes sense to study the Jordan form because you might think it is a corner case which it is because unless you have repeated eigenvalues you never run into trouble see if you have distinct eigenvalues it is always diagonalizable you might think if you have repeated eigenvalues only then this seems to because and this also suggests as much so in general if you have distinct eigenvalues always diagonalizable so why shouldn't we stop our studies there and go further but sometimes the eigenvalues are not just about your numerical computation you see sometimes it is the nature of a physical system that dictates that your eigenvalues will have to be repeated suppose the physical model of a system necessitates that the system will have repeated eigenvalues it is not about your calculation or computation where you can just say oh just give it a little bit of tweaking and perturbation here and there and you immediately have distinct eigenvalues right so it still makes sense so we will now try and establish this any other questions on this okay maybe we will post on this proof to the next module